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Proof step 1 from proof int2bits-rec-2pot-neg-01 (conclusion)

⊦ j ≤ ~(2 ^ i) ∧ 0 < i ∧ j % 2 ^ i = 0 → int2bits(j) = int2bits(j / 2 ^ i) + fillsgn(i→n(i + 1), int2bits(j % 2 ^ i)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j % 2 ^ i = 0, j + 2 ^ i < i, j + 2 ^ i ≤ 0, 0 < i ⊦ int2bits(j) = int2bits(j / 2 ^ i) + mklist(false, i→n(i))

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ (boolvar ' + bits).rest = bits;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ fillsgn(n + 1, boolvar ') = mklist(boolvar, n + 1);% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;